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### Limits of Sequences

```
Date: 02/25/2001 at 02:11:06
From: N. D. Kapoor

Dear Dr. Math,

We know that the limit (n -> infinity) [(1 + 1/n)^n] = e. Is the limit
(n -> infinity) [(1 + 1/sqrt(n))^(1.5n)] also e? Also, what is the
limit (n -> infinity) [(1 + a/n)^n], where a is not equal to 0?

N. D. Kapoor
```

```
Date: 02/26/2001 at 22:27:50
From: Doctor Fenton
Subject: Re: A question about limits

Dear Mr. Kapoor,

Thanks for writing to Dr. Math. For questions like these, I like to
use the fact that for a > 0 and b real,

a^b = e^[b*ln(a)]

where ln is the natural logarithm. Then:

(1 + 1/sqrt(n))^(1.5n) = e^[1.5*n*ln(1 + 1/sqrt(n))]

The exponent can be written as the indeterminate form:

ln( 1 + 1/sqrt(n))
------------------
2/(3n)

By L'Hospital's Rule, the limit of this expression as n->oo is the
same as the limit as n->oo of:

1            1
----------- * ---------
1 + sqrt(n)   2*sqrt(n)
-------------------------
-2
-----
3*n^2

and the quotient is of order n as n->oo, so the expression diverges to
-oo, which means the original expression -> 0, not e.

For the second expression, you can repeat this computation, or write:

(1+a/n)^n = (1 + a/n)^[(n/a)*a]

= [(1 + a/n)^(n/a)]*a

= {[ 1 + 1/(n/a)]^(n/a)}^a

and the term in braces {} approaches e by your earlier observation.

If you have further questions, please write us again.

- Doctor Fenton, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus
High School Sequences, Series

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